Question: $\begin{aligned} &T = -2a^2+a+6 \\\\ &N = -3a^2+2a-5 \end{aligned}$ $N-T=$
Explanation: Since we are asked to find $N-T$, let's substitute in the trinomial expressions that we are given for $N$ and $T$ : $N-T = (-3a^2+2a-5)-(-2a^2+a+6)$ Since we are subtracting, it is helpful to distribute the $\text{{negative sign}}$ across all terms in the second trinomial: $\begin{aligned}&(-3a^2+2a-5){-}(-2a^2+a+6)\\ \\ =&(-3a^2+2a-5){-}(-2a^2){-}a{-}6\\ \\ =&-3a^2+2a-5+2a^2-a-6 \end{aligned}$ Note that the parentheses around the first trinomial don't affect the order of operations, so we can just remove them. When we add or subtract terms in a polynomial expression, the only way that we can simplify the expression is by combining those terms that are alike. Our expression contains terms of $3$ different degrees in the same variable: ${a^2}, {a},$ and the $\text{{constant}}$ term: ${{-3a^2} {+2a} {-5} {+2a^2} {-a} {-6}}$ Note that the negative sign in front of the term ${a}$ means that the coefficient of ${a}$ is ${-1}$. Now that we have identified like terms, let's combine them. Make sure to keep track of positive and negative signs! ${{(-3+2)a^2} + {(2-1)a} + {(-5-6)}}$ When we add the coefficients in front of each term, we get the following trinomial: ${-a^2+a-11}$